Question

我遇到了一个我认为与Function对象（Function.name）上的保留属性发生冲突的问题，我想知道是否有办法让一个静态变量叫做＆＃34;命名＆＃34;在我的家里＆＃34;下面的类，全部为小写，不与Function.name？

冲突

原始javascript：

// Note: I've changed the "home" class to "_c0" here to show the problem more clearly.
var myApi;
(function (myApi) {
    var _c0 = (function () {
        function _c0() { }
        _c0.name = "Home"; // this is colliding with the "Function.name". there is no "name" property on this object during runtime
        _c0.Name = "Home";
        _c0.name1 = "Home";
        _c0.url = "/Home";
        return _c0;
    })();
    myApi.home = _c0;
})(myApi || (myApi = {}));

console.log(myApi.home.name); // prints "_c0" <- this is the name of the function
console.log(myApi.home.Name); // prints "Home"
console.log(myApi.home.name1); // prints "Home"
console.log(myApi.home.url); // prints "/Home"

for(var prop in myApi.home)
    console.log(prop + ": " + myApi.home[prop]);
/* prints: (note the lack of "name: Home" in the output)
    Name: Home
    name1: Home
    url: /Home
*/

打字稿文件：

module myApi {
    export class home {
        static name = "Home";
        static Name = "Home"; // just testing
        static name1 = "Home"; // just testing
        static url = "/Home";
    }
}

console.log(myApi.home.name);
console.log(myApi.home.Name);
console.log(myApi.home.name1);
console.log(myApi.home.url);
for (var prop in myApi.home)
    console.log(prop + ": " + myApi.home[prop]);

Answer 1

没有办法做到这一点。函数对象具有name属性，无法覆盖。可以在下面的代码段中找到简单的证据：

var F = function(){}; 
Object.defineProperty(F, "name", {value: "Home"}); 
//TypeError: Cannot redefine property: name

这意味着静态name属性是只读的。无法输入新值，因为该值的描述符会阻止该值。我建议你为你的房产选择另一个名字。

使用Typescript进行属性名称冲突

1 个答案: